1.已知H2O(l)在373.15K时饱和蒸气压101325Pa,摩尔蒸发焓为40.67kJ·mol-1 试计算
(1)H2O(l)在363.15K时的饱和蒸气压(2)在海拔为4500m的西藏高原上,大气压力只有57300Pa,试计算那里水的沸点。...
(1)H2O(l)在363.15K时的饱和蒸气压
(2)在海拔为4500m的西藏高原上,大气压力只有57300Pa,试计算那里水的沸点。 展开
(2)在海拔为4500m的西藏高原上,大气压力只有57300Pa,试计算那里水的沸点。 展开
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(1) H2O(l) → H2O(g) , △Hmθ = 40.67kJ·mol-1
K1 = p(H2O)1 / pθ = 101325Pa / 100000Pa = 1.01325
ln(K2/K1) = △Hmθ(T2 - T1) / R·T1·T2
= 40.67×1000J·mol-1×(363.15K - 373.15K) / (8.314J·mol-1·K-1×363.15K×373.15K )
= - 0.36171
K2/K1 = 0.69649
K2 = 0.69649 × 1.01325 = 0.70572
p(H2O)2 = K2 × pθ = 0.70572 × 100KPa = 70.572KPa
(2) K3 = p(H2O)3 / pθ = 57300Pa / 100000Pa = 0.573
ln(K3/K1) = ln(0.573 / 1.01325) = - 0.57
ln(K3/K1) = (△Hmθ / R)·(1 / T1 - 1 / T3)
1 / T3 = 1 / T1 - [ ln(K3/K1)×R / △Hmθ ]
= 1 / 373.15K - (- 0.57)×8.314J·mol-1·K-1/40.67×1000J·mol-1
= 0.0028 K-1
T3 = 357.56K = 84.41℃
K1 = p(H2O)1 / pθ = 101325Pa / 100000Pa = 1.01325
ln(K2/K1) = △Hmθ(T2 - T1) / R·T1·T2
= 40.67×1000J·mol-1×(363.15K - 373.15K) / (8.314J·mol-1·K-1×363.15K×373.15K )
= - 0.36171
K2/K1 = 0.69649
K2 = 0.69649 × 1.01325 = 0.70572
p(H2O)2 = K2 × pθ = 0.70572 × 100KPa = 70.572KPa
(2) K3 = p(H2O)3 / pθ = 57300Pa / 100000Pa = 0.573
ln(K3/K1) = ln(0.573 / 1.01325) = - 0.57
ln(K3/K1) = (△Hmθ / R)·(1 / T1 - 1 / T3)
1 / T3 = 1 / T1 - [ ln(K3/K1)×R / △Hmθ ]
= 1 / 373.15K - (- 0.57)×8.314J·mol-1·K-1/40.67×1000J·mol-1
= 0.0028 K-1
T3 = 357.56K = 84.41℃
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